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I'm working through my book, on the section about compound angle formulae. I've been made aware of the identity $\sin(A + B) \equiv \sin A\cos B + \cos A\sin B$. Next task was to replace B with -B to show $\sin(A - B) \equiv \sin A\cos B - \cos A \sin B$ which was fairly easy. I'm struggling with the following though:

"In the identity $\sin(A - B) \equiv \sin A\cos B - \cos A\sin B$, replace A by $(\frac{1}{2}\pi - A)$ to show that $\cos(A + B) \equiv \cos A\cos B - \sin A\sin B$."

I've got $\sin((\frac{\pi}{2} - A) - B) \equiv \cos A\cos B - \sin A\sin B$ by replacing $\sin(\frac{\pi}{2} - A)$ with $\cos A$ and $\cos(\frac{\pi}{2} - A)$ with $\sin A$ on the RHS of the identity. It's just the LHS I'm stuck with and don't know how to manipulate to make it $\cos(A + B)$.

P.S. I know I'm asking assistance on extremely trivial stuff, but I've been staring at this for a while and don't have a tutor so hope someone will help!

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Hint $\sin((\frac{\pi}{2} - A) - B) =\sin((\frac{\pi}{2} - (A + B)) $ and what is $\sin (\frac{\pi}{2}-$angle$)$? – N. S. May 24 '11 at 21:56
please do accept an answer. – user9413 May 25 '11 at 6:18
@Chandru: Answer accepted now. (N.B. I am in the UK and posted this just before going to bed). – PeteUK May 25 '11 at 9:55
Hey no problem.I just reminded you :) – user9413 May 25 '11 at 9:55
@user9176: This is not the first time I've been stuck on a trigonometry problem when the answer is reachable via some trivial algebraic manipulation. Thanks for the comment. – PeteUK May 25 '11 at 9:57
up vote 5 down vote accepted

$$\sin\left(\left(\frac{\pi}{2} - A\right) - B\right) =\sin\left(\frac{\pi}{2} - (A+B) \right)= \cos (A+B)$$

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Very clear and concise answer. Thanks! – PeteUK May 25 '11 at 9:54

Note that you can also establish:

$$\sin\left(\left(\frac{\pi}{2} - A\right) - B\right) =\sin\left(\frac{\pi}{2} - (A + B)\right) = \cos(A+B)$$ by using the second identity you figured out above, $\sin(A - B) \equiv \sin A\cos B - \cos A\sin B$, giving you:

$$\sin\left(\left(\frac{\pi}{2} - A\right) - B\right) = \sin\left(\frac{\pi}{2} - (A+B)\right)$$ $$ = \sin\left(\frac{\pi}{2}\right)\cos(A+B) - \cos\left(\frac{\pi}{2}\right)\sin(A+B)$$ $$= (1)\cos(A+B) - (0)\sin(A+B)$$ $$ = \cos(A+B)$$

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thanks. I enjoyed the extra bit going through the second identity too. – PeteUK May 25 '11 at 9:43

This is the same as Henry's answer, only presented differently.
$\left(\displaystyle \frac{\pi}{2}-A\right) - B = \displaystyle \frac{\pi}{2} - (A+B)$. Now you can use the fact that $\sin \left(\displaystyle \frac{\pi}{2} - C\right) = \cos(C)$...

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Appreciate your answer. – PeteUK May 25 '11 at 9:53

Update: Since $\sin (\frac{\pi }{2}-x)\equiv \cos x$, you have

$$\sin \left(\frac{\pi }{2}-A-B\right)\equiv \sin \left(\frac{\pi }{2}-(A+B)\right)\equiv \cos (A+B).$$

Replacing $A$ by $\frac{\pi }{2}-A$ in

$$\sin (A-B)\equiv \sin A\cos B-\cos A\sin B,$$


$$\sin \left(\frac{\pi }{2}-A-B\right)\equiv \sin \left( \frac{\pi }{2}-A\right) \cos B-\cos \left( \frac{\pi }{2}-A\right) \sin B.$$

Since $\sin (\frac{\pi }{2}-x)\equiv \cos x$ and $\cos (\frac{\pi }{2}-x)\equiv \sin x$, you have

$$\sin \left(\frac{\pi }{2}-A-B\right)\equiv \sin \left(\frac{\pi }{2}-(A+B)\right)\equiv \cos (A+B),$$

$$\sin \left( \frac{\pi }{2}-A\right) \equiv \cos A,$$


$$\cos \left( \frac{\pi }{2}-A\right) \equiv \sin A.$$


$$\cos (A+B)\equiv \cos A\cos B-\sin A\sin B.$$

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érico: thank you very much for helping me out again. – PeteUK May 25 '11 at 9:52

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